3.66/1.79 YES 3.66/1.80 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 3.66/1.80 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.66/1.80 3.66/1.80 3.66/1.80 Left Termination of the query pattern 3.66/1.80 3.66/1.80 app2(a,g,g) 3.66/1.80 3.66/1.80 w.r.t. the given Prolog program could successfully be proven: 3.66/1.80 3.66/1.80 (0) Prolog 3.66/1.80 (1) PrologToPiTRSProof [SOUND, 0 ms] 3.66/1.80 (2) PiTRS 3.66/1.80 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 3.66/1.80 (4) PiDP 3.66/1.80 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 3.66/1.80 (6) PiDP 3.66/1.80 (7) UsableRulesProof [EQUIVALENT, 0 ms] 3.66/1.80 (8) PiDP 3.66/1.80 (9) PiDPToQDPProof [SOUND, 0 ms] 3.66/1.80 (10) QDP 3.66/1.80 (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.66/1.80 (12) YES 3.66/1.80 3.66/1.80 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (0) 3.66/1.80 Obligation: 3.66/1.80 Clauses: 3.66/1.80 3.66/1.80 app1(.(X, Xs), Ys, .(X, Zs)) :- app1(Xs, Ys, Zs). 3.66/1.80 app1([], Ys, Ys). 3.66/1.80 app2(.(X, Xs), Ys, .(X, Zs)) :- app2(Xs, Ys, Zs). 3.66/1.80 app2([], Ys, Ys). 3.66/1.80 3.66/1.80 3.66/1.80 Query: app2(a,g,g) 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (1) PrologToPiTRSProof (SOUND) 3.66/1.80 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 3.66/1.80 3.66/1.80 app2_in_3: (f,b,b) 3.66/1.80 3.66/1.80 Transforming Prolog into the following Term Rewriting System: 3.66/1.80 3.66/1.80 Pi-finite rewrite system: 3.66/1.80 The TRS R consists of the following rules: 3.66/1.80 3.66/1.80 app2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, app2_in_agg(Xs, Ys, Zs)) 3.66/1.80 app2_in_agg([], Ys, Ys) -> app2_out_agg([], Ys, Ys) 3.66/1.80 U2_agg(X, Xs, Ys, Zs, app2_out_agg(Xs, Ys, Zs)) -> app2_out_agg(.(X, Xs), Ys, .(X, Zs)) 3.66/1.80 3.66/1.80 The argument filtering Pi contains the following mapping: 3.66/1.80 app2_in_agg(x1, x2, x3) = app2_in_agg(x2, x3) 3.66/1.80 3.66/1.80 .(x1, x2) = .(x1, x2) 3.66/1.80 3.66/1.80 U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) 3.66/1.80 3.66/1.80 app2_out_agg(x1, x2, x3) = app2_out_agg(x1) 3.66/1.80 3.66/1.80 3.66/1.80 3.66/1.80 3.66/1.80 3.66/1.80 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 3.66/1.80 3.66/1.80 3.66/1.80 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (2) 3.66/1.80 Obligation: 3.66/1.80 Pi-finite rewrite system: 3.66/1.80 The TRS R consists of the following rules: 3.66/1.80 3.66/1.80 app2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, app2_in_agg(Xs, Ys, Zs)) 3.66/1.80 app2_in_agg([], Ys, Ys) -> app2_out_agg([], Ys, Ys) 3.66/1.80 U2_agg(X, Xs, Ys, Zs, app2_out_agg(Xs, Ys, Zs)) -> app2_out_agg(.(X, Xs), Ys, .(X, Zs)) 3.66/1.80 3.66/1.80 The argument filtering Pi contains the following mapping: 3.66/1.80 app2_in_agg(x1, x2, x3) = app2_in_agg(x2, x3) 3.66/1.80 3.66/1.80 .(x1, x2) = .(x1, x2) 3.66/1.80 3.66/1.80 U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) 3.66/1.80 3.66/1.80 app2_out_agg(x1, x2, x3) = app2_out_agg(x1) 3.66/1.80 3.66/1.80 3.66/1.80 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (3) DependencyPairsProof (EQUIVALENT) 3.66/1.80 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 3.66/1.80 Pi DP problem: 3.66/1.80 The TRS P consists of the following rules: 3.66/1.80 3.66/1.80 APP2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> U2_AGG(X, Xs, Ys, Zs, app2_in_agg(Xs, Ys, Zs)) 3.66/1.80 APP2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APP2_IN_AGG(Xs, Ys, Zs) 3.66/1.80 3.66/1.80 The TRS R consists of the following rules: 3.66/1.80 3.66/1.80 app2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, app2_in_agg(Xs, Ys, Zs)) 3.66/1.80 app2_in_agg([], Ys, Ys) -> app2_out_agg([], Ys, Ys) 3.66/1.80 U2_agg(X, Xs, Ys, Zs, app2_out_agg(Xs, Ys, Zs)) -> app2_out_agg(.(X, Xs), Ys, .(X, Zs)) 3.66/1.80 3.66/1.80 The argument filtering Pi contains the following mapping: 3.66/1.80 app2_in_agg(x1, x2, x3) = app2_in_agg(x2, x3) 3.66/1.80 3.66/1.80 .(x1, x2) = .(x1, x2) 3.66/1.80 3.66/1.80 U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) 3.66/1.80 3.66/1.80 app2_out_agg(x1, x2, x3) = app2_out_agg(x1) 3.66/1.80 3.66/1.80 APP2_IN_AGG(x1, x2, x3) = APP2_IN_AGG(x2, x3) 3.66/1.80 3.66/1.80 U2_AGG(x1, x2, x3, x4, x5) = U2_AGG(x1, x5) 3.66/1.80 3.66/1.80 3.66/1.80 We have to consider all (P,R,Pi)-chains 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (4) 3.66/1.80 Obligation: 3.66/1.80 Pi DP problem: 3.66/1.80 The TRS P consists of the following rules: 3.66/1.80 3.66/1.80 APP2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> U2_AGG(X, Xs, Ys, Zs, app2_in_agg(Xs, Ys, Zs)) 3.66/1.80 APP2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APP2_IN_AGG(Xs, Ys, Zs) 3.66/1.80 3.66/1.80 The TRS R consists of the following rules: 3.66/1.80 3.66/1.80 app2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, app2_in_agg(Xs, Ys, Zs)) 3.66/1.80 app2_in_agg([], Ys, Ys) -> app2_out_agg([], Ys, Ys) 3.66/1.80 U2_agg(X, Xs, Ys, Zs, app2_out_agg(Xs, Ys, Zs)) -> app2_out_agg(.(X, Xs), Ys, .(X, Zs)) 3.66/1.80 3.66/1.80 The argument filtering Pi contains the following mapping: 3.66/1.80 app2_in_agg(x1, x2, x3) = app2_in_agg(x2, x3) 3.66/1.80 3.66/1.80 .(x1, x2) = .(x1, x2) 3.66/1.80 3.66/1.80 U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) 3.66/1.80 3.66/1.80 app2_out_agg(x1, x2, x3) = app2_out_agg(x1) 3.66/1.80 3.66/1.80 APP2_IN_AGG(x1, x2, x3) = APP2_IN_AGG(x2, x3) 3.66/1.80 3.66/1.80 U2_AGG(x1, x2, x3, x4, x5) = U2_AGG(x1, x5) 3.66/1.80 3.66/1.80 3.66/1.80 We have to consider all (P,R,Pi)-chains 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (5) DependencyGraphProof (EQUIVALENT) 3.66/1.80 The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node. 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (6) 3.66/1.80 Obligation: 3.66/1.80 Pi DP problem: 3.66/1.80 The TRS P consists of the following rules: 3.66/1.80 3.66/1.80 APP2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APP2_IN_AGG(Xs, Ys, Zs) 3.66/1.80 3.66/1.80 The TRS R consists of the following rules: 3.66/1.80 3.66/1.80 app2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, app2_in_agg(Xs, Ys, Zs)) 3.66/1.80 app2_in_agg([], Ys, Ys) -> app2_out_agg([], Ys, Ys) 3.66/1.80 U2_agg(X, Xs, Ys, Zs, app2_out_agg(Xs, Ys, Zs)) -> app2_out_agg(.(X, Xs), Ys, .(X, Zs)) 3.66/1.80 3.66/1.80 The argument filtering Pi contains the following mapping: 3.66/1.80 app2_in_agg(x1, x2, x3) = app2_in_agg(x2, x3) 3.66/1.80 3.66/1.80 .(x1, x2) = .(x1, x2) 3.66/1.80 3.66/1.80 U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) 3.66/1.80 3.66/1.80 app2_out_agg(x1, x2, x3) = app2_out_agg(x1) 3.66/1.80 3.66/1.80 APP2_IN_AGG(x1, x2, x3) = APP2_IN_AGG(x2, x3) 3.66/1.80 3.66/1.80 3.66/1.80 We have to consider all (P,R,Pi)-chains 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (7) UsableRulesProof (EQUIVALENT) 3.66/1.80 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (8) 3.66/1.80 Obligation: 3.66/1.80 Pi DP problem: 3.66/1.80 The TRS P consists of the following rules: 3.66/1.80 3.66/1.80 APP2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APP2_IN_AGG(Xs, Ys, Zs) 3.66/1.80 3.66/1.80 R is empty. 3.66/1.80 The argument filtering Pi contains the following mapping: 3.66/1.80 .(x1, x2) = .(x1, x2) 3.66/1.80 3.66/1.80 APP2_IN_AGG(x1, x2, x3) = APP2_IN_AGG(x2, x3) 3.66/1.80 3.66/1.80 3.66/1.80 We have to consider all (P,R,Pi)-chains 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (9) PiDPToQDPProof (SOUND) 3.66/1.80 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (10) 3.66/1.80 Obligation: 3.66/1.80 Q DP problem: 3.66/1.80 The TRS P consists of the following rules: 3.66/1.80 3.66/1.80 APP2_IN_AGG(Ys, .(X, Zs)) -> APP2_IN_AGG(Ys, Zs) 3.66/1.80 3.66/1.80 R is empty. 3.66/1.80 Q is empty. 3.66/1.80 We have to consider all (P,Q,R)-chains. 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (11) QDPSizeChangeProof (EQUIVALENT) 3.66/1.80 By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 3.66/1.80 3.66/1.80 From the DPs we obtained the following set of size-change graphs: 3.66/1.80 *APP2_IN_AGG(Ys, .(X, Zs)) -> APP2_IN_AGG(Ys, Zs) 3.66/1.80 The graph contains the following edges 1 >= 1, 2 > 2 3.66/1.80 3.66/1.80 3.66/1.80 ---------------------------------------- 3.66/1.80 3.66/1.80 (12) 3.66/1.80 YES 3.86/1.83 EOF